Answer to Question #98913 in General Chemistry for Sarah G.

Question

An empty steel container is filled with 0.0500 atm of HF. The system is allowed to reach equilibrium. If Kp = 2.76 for the reaction below, what is the equilibrium partial pressure of H₂?

Answer 1

$$2HF = H_2 + F_2$$

Suppose the system initially contained 2 mol HF, then:

$$K_p = {x^2 \over 2 - 2x}$$

$$2.76(2 - 2x) = x^2$$

$$5.52 - 5.52x = x^2$$

$$x^2 + 5.52x - 5.52 = 0$$

$$x = 0.86$$

Then, the amount of substance substances in the reaction in equilibrium:

$$n(HF) = 0.28 mol$$

$$n(H_2) = n(F_2) = 0.86 mol$$

$$\chi(H_2) = {0.86 \over 0.86 + 0.86 + 0.28} = 0.43$$

Then:

$$P_{H_2} = \chi(H_2)*P$$

$$P_{H_2} = 0.43*0.0500atm = 0.0215atm$$

Answer: 0.0215 atm